Markov’s and Chebyshev’s Inequalities

STA 211 · Lecture 06

Introduction

Probability inequalities provide bounds when the full distribution is unknown. Markov’s inequality uses only a non-negative mean; Chebyshev’s inequality uses the mean and variance. Their bounds may be conservative, but they hold for broad classes of distributions.

Learning Objectives

You should be able to:

• state and prove Markov’s inequality;
• derive Chebyshev’s inequality;
• calculate distribution-free probability bounds; and
• interpret the limits of these bounds.

Markov’s Inequality

If $X$ is non-negative and $a > 0$, then

$$P(X \geq a) \leq \frac{E(X)}{a}.$$

Proof

Because $X \geq a$ whenever the event $\{ X \geq a\}$ occurs,

$$X \geq aI_{\{ X \geq a\}},$$

where $I$ is an indicator. Taking expectations gives

$$E(X) \geq aP(X \geq a),$$

and the result follows.

Example 6.1

If the mean monthly claim is ₦40,000, then without assuming a distribution,

$$P(X \geq 100,000) \leq \frac{40,000}{100,000} = 0.4.$$

Chebyshev’s Inequality

If $X$ has mean $\mu$ and finite variance $\sigma^{2}$, then for every $k > 0$,

$$P\left( |X - \mu| \geq k\sigma \right) \leq \frac{1}{k^{2}}.$$

Equivalently,

$$P\left( |X - \mu| < k\sigma \right) \geq 1 - \frac{1}{k^{2}}.$$

Derivation

Apply Markov’s inequality to the non-negative variable $(X - \mu)^{2}$:

$$\begin{matrix} P\left( |X - \mu| \geq k\sigma \right) & = P\left( (X - \mu)^{2} \geq k^{2}\sigma^{2} \right) \\ & \leq \frac{E\left\lbrack (X - \mu)^{2} \right\rbrack}{k^{2}\sigma^{2}} \\ & = \frac{1}{k^{2}}. \end{matrix}$$

Example 6.2: A numerical interval

A distribution has mean 50 and standard deviation 6. At least what proportion lies between 38 and 62?

The interval is $50 \pm 12 = 50 \pm 2(6)$, so $k = 2$:

$$P(38 < X < 62) \geq 1 - \frac{1}{2^{2}} = 0.75.$$

Example 6.3: Required width

How many standard deviations are needed to guarantee at least 90% of observations?

We require

$$1 - \frac{1}{k^{2}} \geq 0.90.$$

Thus $1/k^{2} \leq 0.10$, so

$$k \geq \sqrt{10} \approx 3.162.$$

Chebyshev versus the Empirical Rule

Chebyshev’s inequality applies to any distribution with finite variance. For an approximately normal distribution, the empirical rule gives much tighter descriptive percentages: about 68%, 95% and 99.7% within 1, 2 and 3 standard deviations. These normal percentages should not be used when the shape is unknown.

Post-Test

1. State the assumptions required for Markov’s inequality.
2. A non-negative random variable has mean 12. Find an upper bound for $P(X \geq 48)$.
3. A variable has mean 100 and variance 225. Find a lower bound for $P(70 < X < 130)$.
4. Find $k$ such that Chebyshev guarantees at least 96% within $k$ standard deviations.
5. Explain why Chebyshev’s bound may not equal the actual probability.

Selected answers: 2. $0.25$; 3. $1 - 1/2^{2} = 0.75$; 4. $k \geq 5$.

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